ECE 333 Renewable Energy Systems

Lecture 13: Per Unit, Power Flow

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

overbye@illinois.edu

Announcements

• First exam average: 74.1 • Per unit and power flow is only covered in lecture; not

in the book• No quiz on March 12• HW 6 is posted on the website and is due on Thursday

March 19. HW 6 must be turned in and will count the same as a quiz. There will be no quiz on March 19.

• Start Reading Chapter 4 (The Solar Resource)

2

Wind Speed Doubling and Wind Turbine Generator (WTG) Output

• Just about everyone missed problem 3.d, asking about what happens to the output of a WTG if the wind speed doubles; the answer was "It Depends"– Power in wind goes up by 8, but not usually the power output

3Image: http://site.ge-energy.com/prod_serv/products/wind_turbines/en/downloads/GEA14954C15-MW-Broch.pdf

Image shows power output for GE 1.5 and 1.6 MW WTGs;cut-in speed is 3.5 m/s, while cut-out is 25 m/s

Wind Power and the Power Flow

• The most common power system analysis tool is the power flow (also known sometimes as the load flow)– power flow determines how the power flows in a network– also used to determine all bus voltages and all currents– because of constant power models, power flow is a

nonlinear analysis technique– power flow is a steady-state analysis tool– it can be used as a tool for planning the location of new

generation, including wind

4

Simplified Power System Modeling

• Balanced three phase systems can be analyzed using per phase analysis

• A “per unit” normalization is simplify the analysis of systems with different voltage levels.

• To provide an introduction to power flow analysis we need models for the different system devices:– Transformers and Transmission lines, generators and loads

• Transformers and transmission lines are modeled as a series impedances

5

Load Models

• Ultimate goal is to supply loads with electricity at constant frequency and voltage

• Electrical characteristics of individual loads matter, but usually they can only be estimated– actual loads are constantly changing, consisting of a large

number of individual devices– only limited network observability of load characteristics

• Aggregate models are typically used for analysis• Two common models

– constant power: Si = Pi + jQi

– constant impedance: Si = |V|2 / Zi

6

Generator Models

• Engineering models depend upon application• Generators are usually synchronous machines• For generators we will use two different models:

– a steady-state model, treating the generator as a constant power source operating at a fixed voltage; this model will be used for power flow and economic analysis

– This model works fairly well for type 3 and type 4 wind turbines

– Other models include treating as constant real power with a fixed power factor.

7

Per Unit Calculations

• A key problem in analyzing power systems is the large number of transformers. – It would be very difficult to continually have to refer

impedances to the different sides of the transformers

• This problem is avoided by a normalization of all variables.

• This normalization is known as per unit analysis.

actual quantityquantity in per unit

base value of quantity

8

Per Unit Conversion Procedure, 1f

1. Pick a 1f VA base for the entire system, SB

2. Pick a voltage base for each different voltage level, VB. Voltage bases are related by transformer turns ratios. Voltages are line to neutral.

3. Calculate the impedance base, ZB= (VB)2/SB

4. Calculate the current base, IB = VB/ZB

5. Convert actual values to per unit

Note, per unit conversion on affects magnitudes, not

the angles. Also, per unit quantities no longer have

units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)9

Per Unit Solution Procedure

1. Convert to per unit (p.u.) (many problems are already in per unit)

2. Solve

3. Convert back to actual as necessary

10

Per Unit Example

Solve for the current, load voltage and load power

in the circuit shown below using per unit analysis

with an SB of 100 MVA, and voltage bases of

8 kV, 80 kV and 16 kV.

Original Circuit11

Per Unit Example, cont’d

2

2

2

80.64

100

8064

100

162.56

100

LeftB

MiddleB

RightB

kVZ

MVA

kVZ

MVA

kVZ

MVA

Same circuit, with

values expressed

in per unit.

12

Per Unit Example, cont’d

L

2*

1.0 00.22 30.8 p.u. (not amps)

3.91 2.327

V 1.0 0 0.22 30.8

p.u.

0.189 p.u.

1.0 0 0.22 30.8 30.8 p.u.

LL L L

G

Ij

VS V I

ZS

13

Per Unit Example, cont’d

To convert back to actual values just multiply the

per unit values by their per unit base

LActual

ActualL

ActualG

MiddleB

ActualMiddle

0.859 30.8 16 kV 13.7 30.8 kV

0.189 0 100 MVA 18.9 0 MVA

0.22 30.8 100 MVA 22.0 30.8 MVA

100 MVAI 1250 Amps

80 kV

I 0.22 30.8 Amps 275 30.8

V

S

S

14

Three Phase Per Unit

1. Pick a 3f VA base for the entire system,

2. Pick a voltage base for each different voltage level, VB. Voltages are line to line.

3. Calculate the impedance base

Procedure is very similar to 1f except we use a 3f VA base, and use line to line voltage bases

3BSf

2 2 2, , ,3 1 1

( 3 )

3B LL B LN B LN

BB B B

V V VZ

S S Sf f f

Exactly the same impedance bases as with single phase!15

Three Phase Per Unit, cont'd

4. Calculate the current base, IB

5. Convert actual values to per unit

3 1 13 1B B

, , ,

3I I

3 3 3B B B

B LL B LN B LN

S S S

V V V

f f ff f

Exactly the same current bases as with single phase!

16

Three Phase Per Unit Example

Solve for the current, load voltage and load power in the previous circuit, assuming a 3f power base of300 MVA, and line to line voltage bases of 13.8 kV,138 kV and 27.6 kV (square root of 3 larger than the 1f example voltages). Also assume the generator is Y-connected so its line to line voltage is 13.8 kV.

Convert to per unit

as before. Note the

system is exactly the

same!

17

3f Per Unit Example, cont'd

L

2*

1.0 00.22 30.8 p.u. (not amps)

3.91 2.327

V 1.0 0 0.22 30.8

p.u.

0.189 p.u.

1.0 0 0.22 30.8 30.8 p.u.

LL L L

G

Ij

VS V I

ZS

Again, analysis is exactly the same!

18

3f Per Unit Example, cont'd

LActual

ActualL

ActualG

MiddleB

ActualMiddle

0.859 30.8 27.6 kV 23.8 30.8 kV

0.189 0 300 MVA 56.7 0 MVA

0.22 30.8 300 MVA 66.0 30.8 MVA

300 MVAI 125 (same cur0 Amps

3138 kV

I 0.22 30.

rent!)

8

V

S

S

Amps 275 30.8

Differences appear when we convert back to actual values

19

3f Per Unit Example 2

•Assume a 3f load of 100+j50 MVA with VLL of 69 kV is connected to a source through the below network:

What is the supply current and complex power?

Answer: I=467 amps, S = 103.3 + j76.0 MVA

20

Power Flow Analysis

• We now have the necessary models to start to develop the power system analysis tools

• The most common power system analysis tool is the power flow (also known sometimes as the load flow)– power flow determines how the power flows in a network– also used to determine all bus voltages and all currents– because of constant power models, power flow is a

nonlinear analysis technique– power flow is a steady-state analysis tool

21

Linear Power System Elements

Resistors, inductors, capacitors, independent

voltage sources and current sources are linear

circuit elements

1V = R I V = V =

Such systems may be analyzed by superposition

j L I Ij C

22

Nonlinear Power System Elements

•Constant power loads and generator injections are nonlinear and hence systems with these elements can not be analyzed by superposition

Nonlinear problems can be very difficult to solve,

and usually require an iterative approach

23

Nonlinear Systems May Have Multiple Solutions or No Solution

•Example 1: x2 - 2 = 0 has solutions x = 1.414…•Example 2: x2 + 2 = 0 has no real solution

f(x) = x2 - 2 f(x) = x2 + 2

two solutions where f(x) = 0 no solution f(x) = 0

24

Multiple Solution Example 3

• The dc system shown below has two solutions:

where the 18 watt

load is a resistive

load

22

Load

Load

Load

The equation we're solving is

9 voltsI 18 watts

1 +R

One solution is R 2

Other solution is R 0.5

Load LoadR R

What is the

maximum

PLoad?

25

Adding New Generation

• Adding (siting) large amounts of new generation, such as for a wind farm, often require changes to the transmission system– Developers try to site where existing capacity is available

• Getting new right-of-ways for transmission lines can be quite difficult (i.e., not in my backyard [NIMBY]) even when the transmission is associated with "green" energy

• High voltage (e.g., 345 kV) overhead transmission lines can cost $2 million per mile, whereas burying the conductors can increase costs by a factor of 10– AC lines cannot be buried for long distances because of the

high capacitance associated with underground lines26

Adding New Generation and Transmission

• Power flow studies are done to determine how much new transmission is required, and which would be the best right-of-ways– Often there is no single best answer

• Example: Ameren Illinois Rivers Project

27Project Website: http://www.ilriverstransmission.com/maps

Ameren Illinois Rivers Project: Sidney to Rising 345 kV Line

• One portion of project is adding a new 28 mile 345 kV transmission line in Champaign County – Estimated in-service Nov 2016, cost $66 million

28

Bus Admittance Matrix or Ybus

• First step in solving the power flow is to create what is known as the bus admittance matrix, often call the Ybus.

• The Ybus gives the relationships between all the bus current injections, I, and all the bus voltages, V,

I = Ybus V

• The Ybus is developed by applying KCL at each bus in the system to relate the bus current injections, the bus voltages, and the branch impedances and admittances

29

Ybus Example

Determine the bus admittance matrix for the network

shown below, assuming the current injection at each

bus i is Ii = IGi - IDi where IGi is the current injection into the bus from the generator and IDi is the current flowing into the load

30

Ybus Example, cont’d

1 1 1

1 2 1 31 12 13

1 1 2 1 3 j

1 2 3

2 21 23 24

1 2 3 4

By KCL at bus 1 we have

1( ) ( ) (with Y )

( )

Similarly

( )

G D

A B

A Bj

A B A B

A A C D C D

I I I

V V V VI I I

Z Z

I V V Y V V YZ

Y Y V Y V Y V

I I I I

Y V Y Y Y V Y V Y V

31

Ybus Example, cont’d

1 1

2 2

3 3

4 4

We can get similar relationships for buses 3 and 4

The results can then be expressed in matrix form

0

0

0 0

bus

A B A B

A A C D C D

B C B C

D D

I Y Y Y Y V

I Y Y Y Y Y Y V

I Y Y Y Y V

I Y Y V

I Y V

For a system with n buses, Ybus is an n by n

symmetric matrix (i.e., one where Aij = Aji)

32